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Consider the following expression:

$$ \left[1-\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi\right]^{n-1} $$ where $F:[0,1]\rightarrow [0,1]$ is a continuously differentiable function with $F'=f$, $x\in[0,1]$, and $n>2$. Suppose that $\rho$ belongs to the set of continuous functions defined on $[0,1]$, $C$, and that we use the sup norm in this set. I want to find a function $\rho^*\in C$ such that (with $x<1$ fixed): $$ \left[1-\int_{x}^{1}F(\rho^{*}(\xi))f(\xi)d\xi\right]^{n-1}\geq \left[1-\int_{x}^{1}F(\rho(\xi))f(\xi)d\xi\right]^{n-1} $$ for all $\rho\in C$. Does this make any sense at all? if so, how can I express this?

Thank you so much!

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Forgot to add that F is increasing in all its support... –  Cristian Jul 7 '12 at 4:01
    
Let $I$ be your integral. If $(1-I)^{n-1}$ is maximal, then $1-I$ is maximal, and $I$ is minimal, right? So don't you just need to minimize the integral with respect to $\rho$? –  mjqxxxx Jul 7 '12 at 6:18
    
@mjqxxxx thanks for your comment. You're right (I think). However, the question still is what exactly means to minimize I wrt $\rho$. All that I know is that $\rho$ lives in the $C[0,1]$ space with the sup norm. What I don't understand is the meaning of finding the 'smallest' function that makes the integral minimal...or am I too confused not to see the obvious? Thanks again! –  Cristian Jul 7 '12 at 6:23
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I may be missing something here, but since $F$ is increasing, surely the function $\rho^*(t) = 0$, would satisfy the above equation? –  copper.hat Jul 7 '12 at 6:38
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It's not immediately clear to me that there is a solution then. The possible presence of $F'(\xi)<0$ tweaks the problem a little. I would guess that a solution exists in $L_{\infty}[0,1]$ (set $\rho(t) = \inf_{s \in [0,1]} F(s)$ if $F'(t) \geq 0$, and $\rho(t) = \sup_{s \in [0,1]} F(s)$ otherwise). –  copper.hat Jul 7 '12 at 6:48

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